A truck travelling at 60 km/h on a straight level road shuts off its engine and coasts to a stop. If the frictional forces are 2N per kg of mass and the air resistance is v/3 N per kg mass, how far and how long is it until the truck stops?
Net Force = ma = -(2m + vm/3)
a = -(2 + v/3)
a = -((6+v)/3) = dv/dt
dt/dv = -(3/(6+v))
t = -3ln(6+v) + c
At t = 0, v = 16.6666 m/s
So c = 9.36
t = -3ln(6+v) + 9.36
Let v = 0 to find t
So t = 3.98 seconds
Rearranging this:
t - 9.36 = -3ln(6+v)
(t - 9.36)/-3 = ln(6+v)
6+v = e^((t - 9.36)/-3)
So v = e^((t - 9.36)/-3) - 6
Integrating this from 0 to 3.98 gives 26.03m as the distance until it stops
Is my working right?? It just seems to me that the truck slows down way too fast
Net Force = ma = -(2m + vm/3)
a = -(2 + v/3)
a = -((6+v)/3) = dv/dt
dt/dv = -(3/(6+v))
t = -3ln(6+v) + c
At t = 0, v = 16.6666 m/s
So c = 9.36
t = -3ln(6+v) + 9.36
Let v = 0 to find t
So t = 3.98 seconds
Rearranging this:
t - 9.36 = -3ln(6+v)
(t - 9.36)/-3 = ln(6+v)
6+v = e^((t - 9.36)/-3)
So v = e^((t - 9.36)/-3) - 6
Integrating this from 0 to 3.98 gives 26.03m as the distance until it stops
Is my working right?? It just seems to me that the truck slows down way too fast